Klein-Gordon equation

marzo 31, 2013

Trying to understand the behavior of quantum nature, this was the first attempt to describe the atom. It is known that Schrödinger found this equation, but he did discard it because the equation did not predict the correct spectrum for hydrogen atom.

Consider as a quantum mechanical equation, the Klein-Gordon equation fails miserably describing nature. But consider as a field, the equation open new realms of nature. As a field, the theory describes infinite degrees of freedom, i.e. infinite particles. In this case infinite spin-0 particles. As a quantum mechanical system, the theory leads to paradoxes. But as a field theory is a beautiful building that opens the world of contemporary physics. Let’s look.

As always we begin with a energy-momentum relation. In this case with the relativistic relation.

p^{\mu}p_{\mu}=m^{2}c^{2} .

Now, writting this relation as a relation between operators  p^{\mu}=i\hbar \partial^{\mu} , we find

\left( \partial^{\mu}\partial_{\mu} + m^{2}\right)\phi=0 ,

where  \phi   is the wave function and we have used natural units  c=\hbar=1 . Note that if  m=0 , the Klein-Gordon equation reduces to wave equation

\partial^{\mu}\partial_{\mu}\phi=\dfrac{\partial^{2}\phi}{\partial t^{2}}-\nabla^{2} \phi=0 .

Klein- Gordon equation also admits plane wave solutions. Let’s consider  \phi=e^{\mp ip^{\mu}x_{\mu}} ; it’s like do a Fourier transform. So,

-p^{\mu}p_{\mu} + m^{2} = -E^{2} + \vec{\mathbf{p}}^{2} + m^{2} = 0 ,


E = \pm\sqrt{\vec{\mathbf{p}}^{2} + m^{2}} .

Now remember that we are describing a free particle and we have found that this description allows for both positive and negative energy solutions. This is a little bit awkward. There is nothing bad at all with negative energy; negative energy states describes bound states like planetary motion. But a free particle, a free state, can not has negative energy solutions… at least classically.

Lagrangean Formalism

Consider the Lagrangian density

\mathcal{L} = \partial^{\mu}\phi^{*}\partial_{\mu}\phi - m^{2}\phi^{*}\phi .

Now the  \phi   is a complex field, not a wave function. The equation of motion for this field is

\left( \partial^{\mu}\partial_{\mu} + m^{2}\right)\phi=0 .

Let’s do a transformation to the field, a global phase transformation and consider the conserved current of this theory. The transformation is

\phi' = e^{i\alpha}\phi .

Clearly, the new Lagrangian  \mathcal{L'}   is invariant and also the action  \mathcal{S} ,

\mathcal{L'} = \partial^{\mu}\left( e^{-i\alpha}\phi^{*}\right)\partial_{\mu}\left( e^{i\alpha}\phi\right) - m^{2}\left( e^{-i\alpha}\phi^{*}\right)\left( e^{i\alpha}\phi\right) .


\mathcal{L'} = \partial^{\mu}\phi^{*}\partial_{\mu}\phi - m^{2}\phi^{*}\phi = \mathcal{L} .

Now that we have seen the Lagrangian is invariant, we can write the conserved current as

j^{\mu} = \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_{i})}\delta\phi_{i} ,

where the index i stand for the two fields \phi and its complex conjugate. In order to write the current we also need the infinitesimal field variation  \delta\phi .  That is

\phi' = e^{i\alpha}\phi \hspace{0.5cm} \Longrightarrow \hspace{0.5cm} \phi' = (1 + i\alpha)\phi .

Hence,  \delta\phi = i\alpha\phi .  And the conserved current is

j^{\mu} = \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta\phi + \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi^{*})}\delta\phi^{*} .

Substituting all the pieces,

j^{\mu} = -i\left[\phi^{*}\partial^{\mu}\phi - \phi\partial^{\mu}\phi^{*}\right] .

And taking the divergence and using the equation of motions we verified that

\partial_{\mu}j^{\mu} = 0

For now let’s leave this up to this point here. I’ll see you soon.



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